EDA and predictive modelling of airbnb prices

What's in this notebook:

Data prep and EDA

• summary statistics for each variable
• distribution of prices
• relationship between prices and other variables

Predictive modelling

• Linear regression, with grid search and cross validation
• Decision tree regression, with grid search and cross validation

About the data

I used a data set that contains Airbnb listings in 10 European cities, with the following attributes available for each listing:

• realSum: the full price of accommodation for two people and two nights in EUR
• room_type: the type of the accommodation
• room_shared: dummy variable for shared rooms
• room_private: dummy variable for private rooms
• person_capacity: the maximum number of guests
• host_is_superhost: dummy variable for superhost status
• multi: dummy variable if the listing belongs to hosts with 2-4 offers
• biz: dummy variable if the listing belongs to hosts with more than 4 offers
• cleanliness_rating: cleanliness rating
• guest_satisfaction_overall: overall rating of the listing
• bedrooms: number of bedrooms (0 for studios)
• dist: distance from city centre in km
• metro_dist: distance from nearest metro station in km
• attr_index: attraction index of the listing location
• attr_index_norm: normalised attraction index (0-100)
• rest_index: restaurant index of the listing location
• attr_index_norm: normalised restaurant index (0-100)
• lng & lat: longitude and latitude of the listing location

Data source: Kaggle https://www.kaggle.com/datasets/thedevastator/airbnb-prices-in-european-cities

import pandas as pd
import numpy as np
import plotly.express as px
import plotly.graph_objects as go
import statsmodels.api as sm
from scipy.stats import pearsonr

Data preparation

Import data

cities_list = ['amsterdam', 'athens', 'barcelona', 'berlin', 'budapest', 
            'lisbon', 'london', 'paris', 'rome', 'vienna']

weekdays_dict = {}
weekends_dict = {}

for city in cities_list:
    df_weekdays = pd.read_csv(f'data/{city}_weekdays.csv')
    df_weekends = pd.read_csv(f'data/{city}_weekends.csv')
    
    # city variable
    df_weekdays['city'] = city
    df_weekends['city'] = city
    
    # dummy variable for weekend
    df_weekdays['is_weekend'] = 0
    df_weekends['is_weekend'] = 1
    
    df_weekdays['week_part'] = 'weekday'
    df_weekends['week_part'] = 'weekend'
    
    weekdays_dict[city] = df_weekdays
    weekends_dict[city] = df_weekends
    print(city, f'weekdays: {len(df_weekdays)}, weekends: {len(df_weekends)}')  

#  combine into one dataframe
df = pd.concat(list(weekdays_dict.values())+list(weekends_dict.values())).drop(columns=['Unnamed: 0'])

# convert realSum to price per night
df['price_per_night'] = df['realSum']/2

df.info()

Exploratory data analysis

Athens have the best guest satisfaction and cleanliness ratings, and also the cheapest despite having the highest person capacity.
Amsterdam is the most expensive, presumably because it has the highest average number of bedrooms.
London has the best attraction index but worst guest satisfaction and cleanliness ratings.
Listing prices on weekdays and weekends are similar, except in Amsterdam where prices are higher on weekends.
Listing prices for superhosts and non-superhosts do not differ much in most cities.

averages

df_ = df.groupby('city').agg({'price_per_night':'median', 'room_shared':'mean', 'bedrooms':'mean', 'person_capacity':'mean', 
                              'host_is_superhost':'mean', 'multi':'mean', 'biz':'mean', 
                              'guest_satisfaction_overall':'mean', 'cleanliness_rating':'mean', 
                              'dist':'mean', 'metro_dist':'mean', 
                             'attr_index_norm':'mean', 'rest_index_norm':'mean'}).reset_index().sort_values('price_per_night')
df_

	city	price_per_night	room_shared	bedrooms	person_capacity	host_is_superhost	multi	biz	guest_satisfaction_overall	cleanliness_rating	dist	metro_dist	attr_index_norm	rest_index_norm
1	athens	63.857709	0.002083	1.271402	3.698106	0.428598	0.267424	0.379924	95.003598	9.638447	1.803080	0.478656	5.740839	9.954268
4	budapest	76.491047	0.003481	1.105669	3.540776	0.378916	0.303332	0.348831	94.585281	9.477374	1.872763	0.544059	12.675248	34.529089
8	rome	91.295911	0.001329	1.229755	3.357372	0.326687	0.385953	0.334109	93.122300	9.514678	3.026982	0.819794	10.426968	25.078056
3	berlin	95.587548	0.029388	1.070451	2.774960	0.257246	0.276167	0.174718	94.323671	9.461755	5.257093	0.836064	16.803111	30.666967
2	barcelona	104.149696	0.004236	1.161313	2.616661	0.181433	0.385104	0.325450	91.109072	9.291564	2.116982	0.441248	16.636220	19.376528
9	vienna	104.247014	0.004524	1.102347	3.350297	0.284139	0.279050	0.339836	93.731128	9.472434	3.139488	0.526670	8.762474	4.239580
5	lisbon	112.687617	0.012841	1.272428	3.343398	0.213951	0.239459	0.587541	91.093875	9.370640	1.966893	0.711482	7.324730	28.274084
6	london	130.647475	0.005004	1.128790	2.846192	0.157410	0.274992	0.387872	90.645652	9.175023	5.326421	1.005547	20.537398	11.234105
7	paris	158.798583	0.014055	0.972787	2.953648	0.140700	0.219498	0.245813	92.037530	9.263606	2.995823	0.227323	18.204358	42.589111
0	amsterdam	230.122091	0.004808	1.292308	2.781731	0.284135	0.283173	0.105288	94.514423	9.465865	2.825052	1.089367	14.246499	26.097566

standard deviation

df_ = df.groupby('city').agg({'price_per_night':'std', 'room_shared':'std', 
                              'bedrooms':'std', 'person_capacity':'std', 
                              'host_is_superhost':'std',  'multi':'std', 'biz':'std',
                              'guest_satisfaction_overall':'std', 'cleanliness_rating':'std', 
                              'dist':'std', 'metro_dist':'std',
                              'attr_index_norm':'std', 'rest_index_norm':'std'}).reset_index().sort_values('price_per_night')
df_

	city	price_per_night	room_shared	bedrooms	person_capacity	host_is_superhost	multi	biz	guest_satisfaction_overall	cleanliness_rating	dist	metro_dist	attr_index_norm	rest_index_norm
5	lisbon	54.486540	0.112596	0.728539	1.344214	0.410128	0.426790	0.492320	9.148114	0.924080	1.742681	0.920204	5.082390	17.877309
8	rome	59.309052	0.036438	0.549710	1.309052	0.469028	0.486847	0.471704	7.815107	0.808415	1.644095	0.631361	6.631054	13.414188
4	budapest	65.572403	0.058903	0.663484	1.256548	0.485177	0.459754	0.476660	6.525680	0.842693	1.874925	0.856410	6.672535	19.111435
3	berlin	117.664645	0.168926	0.552033	1.188142	0.437204	0.447191	0.379802	6.809406	0.849384	3.692649	1.267283	10.774273	16.634505
1	athens	132.940027	0.045600	0.652575	1.284703	0.494922	0.442657	0.485414	8.348637	0.839767	0.953738	0.284154	4.667181	10.778060
7	paris	165.474872	0.117727	0.642571	1.215007	0.347738	0.413937	0.430601	8.818201	0.974036	1.463542	0.122769	7.759372	15.680438
2	barcelona	177.733944	0.064956	0.517108	1.153124	0.385445	0.486706	0.468625	8.607153	1.014577	1.377859	0.284540	9.591798	10.256285
9	vienna	198.873582	0.067115	0.602819	1.282163	0.451067	0.448596	0.473720	7.220808	0.855439	1.942337	0.516132	6.259531	3.683256
0	amsterdam	215.329203	0.069187	0.736683	1.032634	0.451110	0.450648	0.306999	6.350874	0.813421	2.082573	0.831669	10.335158	17.720931
6	london	235.678632	0.070562	0.579477	1.246235	0.364205	0.446533	0.487289	11.510622	1.166180	2.712573	1.263926	11.914575	6.963803

metric_list = ['price_per_night','room_shared','bedrooms','person_capacity',
               'host_is_superhost','multi', 'biz', 'guest_satisfaction_overall','cleanliness_rating',
               'dist', 'metro_dist',  'attr_index_norm', 'rest_index_norm']

highest = []
lowest = []
for metric in metric_list:
    highest.append(df_.sort_values(metric)['city'].values[-1])
    lowest.append(df_.sort_values(metric)['city'].values[0])
    
pd.DataFrame({'metric':metric_list, 'highest':highest, 'lowest':lowest})

	metric	highest	lowest
0	price_per_night	london	lisbon
1	room_shared	berlin	rome
2	bedrooms	amsterdam	barcelona
3	person_capacity	lisbon	amsterdam
4	host_is_superhost	athens	paris
5	multi	rome	paris
6	biz	lisbon	amsterdam
7	guest_satisfaction_overall	london	amsterdam
8	cleanliness_rating	london	rome
9	dist	berlin	athens
10	metro_dist	berlin	paris
11	attr_index_norm	london	athens
12	rest_index_norm	budapest	vienna

fig = go.Figure()

for week_part in ['weekday', 'weekend']:
    fig.add_trace(go.Box(y=df[df['week_part']==week_part]['price_per_night'], 
                         x=df[df['week_part']==week_part]['city'],
                         #boxpoints=False, 
                         name=week_part))# = px.box(df, x='city', y='realSum', color='week_part')

fig.update_layout(
    boxmode='group',
    font_color="black",
    title_font_color="black",
    title='Box plot of listing prices<sup><br>Listing prices on weekdays and weekends are similar, except in Amsterdam where prices are higher on weekends. </sup>', 
    plot_bgcolor='rgba(0, 0, 0, 0)',
    yaxis=dict(title='price per night for two (euros)', 
               range=[0,800], 
               tickformat=',.0f'),
                  

)    
fig.show()

Correlation analysis

Here I computed the Pearson correlation coefficients* between all variables. The following variables to be positively correlated:

• guest satisfaction score & cleanliness rating & being a super host
• price per night & person capacity & number of bedrooms
• distance from city centre & distance from metro
• restaurant and attraction index

These correlations are statistically significant, as shown in p-value matrix.

Nothing surprising here.

^{* Pearson correlation coefficient is the ratio between the covariance of the two variables and the product of their standard deviations. So it is essentially a normalized measurement of covariance.}

vars_of_interest = ['price_per_night','room_shared','bedrooms','person_capacity',
                    'host_is_superhost', 'multi', 'biz',
                    'guest_satisfaction_overall','cleanliness_rating',
                    'dist', 'metro_dist', 'attr_index_norm', 'rest_index_norm']
df_corr = df[vars_of_interest].corr().round(2)

mask = np.triu(np.ones_like(df_corr, dtype=bool))
df_corr_vis = df_corr.mask(mask)

fig = px.imshow(df_corr_vis, text_auto=True, 
               color_continuous_scale=px.colors.diverging.PRGn, 
               color_continuous_midpoint=0)
fig.update_layout(plot_bgcolor='rgba(0, 0, 0, 0)',
                  title='Pearson correlation coefficients',
                  xaxis=dict(tickangle=90), 
                  height=600
                 )

fig.show()

df_pvals = pd.DataFrame(index=vars_of_interest)
for col_metric in vars_of_interest:
    pvals_list = []
    for row_metric in vars_of_interest:
        pvals_list.append(pearsonr(df[col_metric],df[row_metric])[1].round(2))
    df_pvals[col_metric] = pvals_list

df_pvals_vis = df_pvals.mask(mask)

fig = px.imshow(df_pvals_vis, text_auto=True, 
               color_continuous_scale=px.colors.sequential.Purples_r, 
               color_continuous_midpoint=0.1
               )
fig.update_layout(plot_bgcolor='rgba(0, 0, 0, 0)',
                  title='''p-values of the Pearson correlation coefficients
                  <br><sup>(zeros mean that the correlations is so statistically significant that the p-value is smaller than the smallest possible floating point)</sup>''',
                  xaxis=dict(tickangle=90), 
                  height=600
                 )

fig.show()

Fitting OLS trendlines for key variables by city

Scatterplots are a quick way of displaying relationship between two variables. To explore whether prices in different cities might be affected by different variables, I created a scatterplot per city and found that:

• While distance negatively correlates with prices for all cities, it is particularly strong for Amsterdam
• Guest satisfaction is correlated with prices in Amsterdam, Athens, Lisbon, Paris, and Rome, but not in other cities.

fig = px.scatter(df, 
                x='dist', 
                y='price_per_night',
                hover_data=['price_per_night', 'dist'],
                color='city',
                facet_col='city', 
                facet_col_wrap=5,
                title='Distance from centre avs price',
                trendline='ols',
                height=600, width=1000
                )

update_annotations_and_layout(fig)
fig.show()

fig = px.scatter(df, 
                x='metro_dist', 
                y='price_per_night',
                hover_data=['price_per_night', 'metro_dist'],
                color='city',
                facet_col='city', 
                facet_col_wrap=5,
                title='Distance from metro vs price',
                trendline='ols',
                height=600, width=1000
                )

update_annotations_and_layout(fig)
fig.show()

fig = px.scatter(df,#[df['person_capacity']==2], 
                x='guest_satisfaction_overall', 
                y='price_per_night',
                hover_data=['price_per_night', 'guest_satisfaction_overall'],
                color='city',
                facet_col='city', 
                facet_col_wrap=5,
                title='Guest satisfaction vs price',
                trendline='ols',
                height=600, width=1000
                )

update_annotations_and_layout(fig)
fig.show()

Predictive modelling for prices

Here I experimented with two different approaches to predict listing prices, linear regression and decision tree regression. Cross validation is used to identify the best model in each approach. Linear regression performed better than decision tree method, suggesting that the relationship between prices and the explanatory variables are more linear than nonlinear. It is also possible that the decision tree might be overfitting the data, capturing noise and irrelevant patterns instead of the true underlying relationship between variables. Decision trees are more prone to overfitting than linear regression model.

Data preprocessing

# create dummy variables for cities
city_dummies = pd.get_dummies(df['city'], prefix='city')
# create dummy variables for room type
room_type_dummies = pd.get_dummies(df['room_type'], prefix='room_type')

# merge city and room type dummies with the data
df_all_features = pd.concat([df, city_dummies, room_type_dummies], axis=1)

# transform binary categorical variables into dummy variables
for col in ['room_shared', 'host_is_superhost']:
    df_all_features[col] = 1*df_all_features[col]
    
# select the relevant variables and load into x and y
X = df_all_features[['person_capacity', 'bedrooms', 'host_is_superhost', 'biz', 
                     'cleanliness_rating', 'guest_satisfaction_overall', 
                     'dist', 'metro_dist', 'attr_index_norm', 'rest_index_norm', 'is_weekend', 
                    ]+city_dummies.columns.tolist()+room_type_dummies.columns.tolist()].drop(columns=['city_lisbon', 'room_type_Shared room'])

y = np.array(df_all_features['price_per_night']).astype(float)

Standardize features

Here I use StandardScaler() to standardize features to have a mean of zero and a standard deviation of one. This is because I am going to use recursive feature elimination (RFE) for feature selection. RFE ranks features based on the absolute values of the coefficients, therefore it is important that the features are on the same scale.

from sklearn import preprocessing

standard_scaler = preprocessing.StandardScaler()
X = pd.DataFrame(standard_scaler.fit_transform(X), columns=X.columns)

Split the data into training and test sets

from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=42)

Linear Regression

with hyperparameter tuning using grid search cross-validation, evaluated by R-squared (coefficient of determination).

In the script below I will use grid search to tune the number of features to include in the model. The best parameter is identified by cross validation.

from sklearn.linear_model import LinearRegression
from sklearn.model_selection import train_test_split, GridSearchCV, cross_val_score
from sklearn.feature_selection import RFE
from sklearn.metrics import mean_squared_error, r2_score

Prepare data and create instance of grid search object with cross validation

# create an instance of the linear regression model that uses RFE (recursive feature elimination)
lr = LinearRegression()
rfelr = RFE(lr)

# specify the number of features as the hyperparameter to tune
param_grid = [{'n_features_to_select':range(1,len(X.columns))}
               ]
# create an instance of the grid search object, specifying a 5-fold cross validation
grid_search = GridSearchCV(estimator=rfelr, param_grid=param_grid, scoring='r2', cv=5)

Fit the grid search object on the training data

# fit the grid search object on the training data
import time
start_time = time.time()
grid_search.fit(X_train, y_train)
print(f'{(time.time() - start_time):.0f} seconds' )

print('Best hyperparameters: ', grid_search.best_params_)

Fit the model on the training set using the best n_features_to_select and test performance

# fit the model on the training set using the best hyperparameters
lr_best = RFE(LinearRegression(), **grid_search.best_params_)
lr_best = lr_best.fit(X_train, y_train)

# test performance
y_pred = lr_best.predict(X_test)
mse = mean_squared_error(y_test, y_pred)
r2 = r2_score(y_test, y_pred)

print(f'Mean squared error: {mse:.2f}')
print(f'R-squared: {r2:.2f}')

Perform cross validation to get a more robust estimate of the model performance

# perform cross-validation to get a more robust estimate of the model performance

cv_scores = cross_val_score(lr_best, X, y, cv=5, scoring='r2')
print(f'Cross-validation R-squared: {["%.2f" % scr for scr in cv_scores]}' )
print(f'Mean cross-validation R-squared: {cv_scores.mean():.2f}', )

cv_scores = cross_val_score(lr_best, X, y, cv=5, scoring='neg_mean_squared_error')
print(f'Cross-validation neg MSE: {["%.0f" % scr for scr in cv_scores]}')
print(f'Mean cross-validation neg MSE: {cv_scores.mean():.0f}')

Decision tree regressor

An alternative to linear regression is decision tree regression. Here I will be replicating the above process but instead of the number of features, I will be tuning the maximum tree depth instead. Tree depth is the number of splits a tree can make before reaching a decision.

from sklearn.tree import DecisionTreeRegressor

Create instance of grid search object with cross validation

# create an instance of the decision tree regressor
dcr = DecisionTreeRegressor()

# specify maximum depth as the hyperparameter to tune
param_grid = [{'max_depth':range(1,8)}]

# create an instance of grid search object for the decision tree regressor
grid_search_dcr = GridSearchCV(estimator=dcr, scoring='r2', param_grid=param_grid, cv=5)

Fit the grid search object on the training data

start_time = time.time()
# fit the grid search object on the training data
grid_search_dcr.fit(X_train, y_train)
print(f'{(time.time() - start_time):.0f} seconds' )

print('Best hyperparameters: ', grid_search_dcr.best_params_)

Fit the model on the training set using the best max_depth and test performance

# fit the model on the training set using the best hyperparameters
dcr_best = DecisionTreeRegressor(**grid_search_dcr.best_params_)
dcr_best.fit(X_train, y_train)

# test performance
y_pred_dcr = dcr_best.predict(X_test)
mse = mean_squared_error(y_test, y_pred_dcr)
r2 = r2_score(y_test, y_pred_dcr)

print('Mean squared error: ', mse)
print('R-squared: ', r2)

Perform cross validation to get a more robust estimate of the model performance

# perform cross-validation to get a more robust estimate of the model performance

cv_scores = cross_val_score(dcr_best, X, y, cv=5, scoring='r2')
print(f'Cross-validation R-squared: {["%.2f" % scr for scr in cv_scores]}' )
print(f'Mean cross-validation R-squared: {cv_scores.mean():.2f}', )

cv_scores = cross_val_score(dcr_best, X, y, cv=5, scoring='neg_mean_squared_error')
print(f'Cross-validation neg MSE: {["%.0f" % scr for scr in cv_scores]}')
print(f'Mean cross-validation neg MSE: {cv_scores.mean():.0f}')

Statistical inference

RFE picked 17 out of 22 input features for the linear regression model, all 17 features were statistically significant (p-value < 0.05). Here are some interesting observations:

• Restaurant index was eliminated while attraction index was kept. This shows that when deciding where to stay, airbnb guests value proximity to attractions more than restaurants.
• Distance from centre was eliminated while distance from metro remained in the model. Remember from the exploratory data analysis we saw that distance from centre and distance from metro were both negatively correlated with listing price? It turns out that metro distance matters more to airbnb guests.
• The weekend and Vienna dummy variables were also dropped, suggesting that listing prices do not depend on day of the week and that listing prices in Vienna are similar to the 'default' city, Lisbon, which was omitted to avoid multicollinearity.

Using the statsmodel library to retrieve the p-values of each included feature

The decision tree regressor assigned a feature importance score of 0 to all but 6 features had importance score of zero.

Linear regression feature coefficients

pd.DataFrame({'feature_name':lr_best.get_feature_names_out(), 'feature_coef': lr_best.estimator_.coef_}).sort_values('feature_coef', ascending=False)

	feature_name	feature_coef
15	room_type_Entire home/apt	47.581816
6	attr_index_norm	33.434194
7	city_amsterdam	32.171210
1	bedrooms	27.833197
16	room_type_Private room	19.500774
12	city_london	18.830633
13	city_paris	18.778153
0	person_capacity	15.552854
9	city_barcelona	9.085896
2	biz	6.633180
3	cleanliness_rating	2.902304
4	guest_satisfaction_overall	2.836171
10	city_berlin	2.340874
5	metro_dist	-3.746838
14	city_rome	-7.344834
11	city_budapest	-14.770777
8	city_athens	-17.213019

Eliminated features

[feat for feat in lr_best.feature_names_in_ if feat not in lr_best.get_feature_names_out()]

['host_is_superhost', 'dist', 'rest_index_norm', 'is_weekend', 'city_vienna']

Using statsmodel library to get p-values of the feature coefficients

x = df_all_features[lr_best.get_feature_names_out().tolist()]
x = sm.add_constant(x)
model = sm.OLS(y, x).fit()
model.summary()

OLS Regression Results

Dep. Variable:	y	R-squared:	0.241
Model:	OLS	Adj. R-squared:	0.241
Method:	Least Squares	F-statistic:	967.5
Date:	Fri, 17 Mar 2023	Prob (F-statistic):	0.00
Time:	09:29:26	Log-Likelihood:	-3.2992e+05
No. Observations:	51707	AIC:	6.599e+05
Df Residuals:	51689	BIC:	6.600e+05
Df Model:	17
Covariance Type:	nonrobust

	coef	std err	t	P>|t|	[0.025	0.975]
const	-147.9687	10.621	-13.931	0.000	-168.787	-127.151
person_capacity	12.0946	0.691	17.494	0.000	10.739	13.450
bedrooms	43.9366	1.251	35.123	0.000	41.485	46.388
biz	14.0294	1.411	9.940	0.000	11.263	16.796
cleanliness_rating	3.1170	0.947	3.293	0.001	1.262	4.972
guest_satisfaction_overall	0.3429	0.103	3.336	0.001	0.141	0.544
metro_dist	-4.6670	0.812	-5.748	0.000	-6.258	-3.076
attr_index_norm	3.2885	0.081	40.547	0.000	3.130	3.447
city_amsterdam	165.1652	3.610	45.755	0.000	158.090	172.240
city_athens	-57.2870	2.507	-22.852	0.000	-62.201	-52.373
city_barcelona	41.1882	3.272	12.588	0.000	34.775	47.601
city_berlin	12.0632	3.418	3.529	0.000	5.363	18.763
city_budapest	-55.3180	2.752	-20.102	0.000	-60.712	-49.924
city_london	49.3805	2.434	20.289	0.000	44.610	54.151
city_paris	57.1948	2.488	22.993	0.000	52.319	62.070
city_rome	-18.9585	2.153	-8.806	0.000	-23.178	-14.739
room_type_Entire home/apt	100.5021	7.560	13.293	0.000	85.684	115.320
room_type_Private room	42.6894	7.641	5.587	0.000	27.713	57.666

Omnibus:	134639.178	Durbin-Watson:	1.908
Prob(Omnibus):	0.000	Jarque-Bera (JB):	4185074797.675
Skew:	30.258	Prob(JB):	0.00
Kurtosis:	1395.428	Cond. No.	2.09e+03

Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
[2] The condition number is large, 2.09e+03. This might indicate that there are
strong multicollinearity or other numerical problems.

Decision tree regressor feature importances

pd.DataFrame({'feature_name':dcr_best.feature_names_in_, 'feature_importance': dcr_best.feature_importances_}).sort_values('feature_importance', ascending=False)

	feature_name	feature_importance
1	bedrooms	0.404788
8	attr_index_norm	0.388031
11	city_amsterdam	0.093544
0	person_capacity	0.054500
20	room_type_Entire home/apt	0.041960
15	city_budapest	0.017177
4	cleanliness_rating	0.000000
5	guest_satisfaction_overall	0.000000
19	city_vienna	0.000000
18	city_rome	0.000000
17	city_paris	0.000000
16	city_london	0.000000
14	city_berlin	0.000000
13	city_barcelona	0.000000
12	city_athens	0.000000
2	host_is_superhost	0.000000
10	is_weekend	0.000000
9	rest_index_norm	0.000000
3	biz	0.000000
7	metro_dist	0.000000
6	dist	0.000000
21	room_type_Private room	0.000000